1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider the following reactions :
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Hydrocarbons Question 108 English
'A' is :
A
$$CH \equiv CH$$
B
$$C{H_3} - C \equiv C - C{H_3}$$
C
$$C{H_3} - C \equiv CH$$
D
$$C{H_2} = C{H_2}$$
2
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is :
(R = 0.08206 L atm K–1 mol–1)
A
8.2 atm
B
2.46 atm
C
4.92 atm
D
1.64 atm
3
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is :
A
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Haloalkanes and Haloarenes Question 126 English Option 1
B
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Haloalkanes and Haloarenes Question 126 English Option 2
C
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Haloalkanes and Haloarenes Question 126 English Option 3
D
JEE Main 2019 (Online) 12th April Evening Slot Chemistry - Haloalkanes and Haloarenes Question 126 English Option 4
4
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5(g) $$ \to $$ 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :
A
2.083 × 10–3 mol L–1 min–1
B
8.333 × 10–3 mol L–1 min–1
C
4.167 × 10–3 mol L–1 min–1
D
1.667 × 10–2 mol L–1 min–1
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